Field of fractions of $O_L$ where $O_L$ is the integral closure of a integral domain $A$ in a field extension $L$ I have the following problem:

Let $A$ be an integral domain and $K$ its field of fractions. Let $L\supseteq K$ a field that is a finite extension of $K$. Let $O_L$ be the integral closure of $A$ in $L$.
Prove that $L$ is the field of fractions of $O_L$ and that exists a base in $L$ (as $K$-vector space) formed with elements of $O_L$.

I think I solved the first part,
since $L$ is a finite extension i have that $[L:K] = n$ so there is a base with $n$ elements ($L$ is a K-vector space). Then given any $l\in L$, a set of $a_i \in K$, $  a_0 + a_1l+ \cdots + a_nl^n  $ is linearly dipendent so
$$a_0 + a_1l+ \cdots + a_nl^n = 0$$
with $a_0,\ldots,a_n$ not all equal to $0$. This means that $l$ is algebric over $K$ and since $K$ is a field $l$ is also integral over $K$ (just multiply by $\frac{1}{a_n}$ and you get a monic polinomial in $K[x]$).
So now every coefficient is a fraction with numerator and denominator from A. I bring all the fractions to a common denominator $a\not = 0$ and after renaming the coefficients:
$$ l^n + \frac{b_{n-1}}{a}l^{n-1} + \cdots + \frac{b_{1}}{a}l^{} + \frac{b_0}{a} = 0$$
Multiplying by $a^n$ i get
$$ (al)^n + b_{n-1}(al)^{n-1} + \cdots  + b_0 a_{n-1}= 0$$ so $al$ is integral over A because $a_i,b_i \in A$. This means $l = \frac{al}{a}$ where $al,a \in O_L$.
Is this enough? I feel like I am still missing something...
For the second part, the idea is that given $l \in O_L$ then $l$ is a solution for a monic polinomial in $A[x]$, but also in $K[x]$ because every coefficient in $A$ is also in $K$. This means that
$$l^n = - (a_{n-1}l^{n-1} + \cdots + a_0)$$ so that $(1,l,\ldots,l^{n-1})$ is a base and $l \in O_L$.
One last thing, is it possible to prove the first part by saying that: since $L$ is a field and $L \supset O_L$ then $L \supseteq Frac(O_L)$ (where $Frac(O_L)$ is the field of fractions). Then showing that for every field $F$ containing $O_L$ then $F\supseteq L$, and this is always true because every $l\in L$ is generated by elements in $O_L \subset F$.
I'm not expecting anything, if you feel like helping i will appreciate it very much!
 A: There is a nice trick for your second question. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{s}$ be a basis for $L$ over $K$. Since $L/K$ is finite, each $\alpha_{i}$ is algebraic over $K$, and hence satisfies a polynomial equation $f_{i}(X) = 0$ for some $f_{i}(X) \in K[X]$ (monic if you like). Focusing attention on $\alpha_{1}$ (the argument for all the other basis elements will be the same), say $f_{1}(X) = X^{n}+u_{1}X^{n-1}+\cdots+u_{n-1}X+u_{n}$, where $u_{1}, \ldots, u_{n} \in K$. By "clearing denominators'', we obtain a polynomial $g_{1}(X) = a_{0}X^{n}+a_{1}X^{n-1}+\cdots+a_{n} \in A[X]$ (not necessarily monic now, but $a_{0} \neq 0$) such that $g_{1}(\alpha_{1}) = 0$. Multiplying each of the coefficients of $g_{1}(X)$ by $a_{0}^{n-1}$, we see $\alpha_{1}$ is likewise a root of $h_{1}(X) = a_{0}^{n}X^{n}+a_{0}^{n-1}a_{1}X^{n-1}+\cdots+a_{0}^{n-1}a_{n}$. But now notice that $a_{0}\alpha_{1}$ is a root of the monic polynomial $Y^{n}+a_{1}Y^{n-1}+a_{0}a_{2}Y^{n-2}+\cdots+a_{0}^{n-1}a_{n} \in A[X]$, and therefore is integral over $A$. In this way, there is an element $w_{i} \in A$ such that $w_{i}\alpha_{i} \in \mathcal{O}_{L}$. Since scaling our basis by elements of $A$ doesn't change its $K$-span (or affect $K$-linear independence), our desired basis is $w_{1}\alpha_{1}, \ldots, w_{s}\alpha_{s}$.   
